IRF530 Push Pull Amplifier - Part 4

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  • Опубликовано: 25 окт 2024

Комментарии • 38

  • @azav8raa
    @azav8raa 3 года назад +1

    Hi Ian, good stuff and I have enjoyed the entire video series. One thing to look at: On the schematic in part 1, it shows a 1:16 (impedance ratio) for the final transformer. The one you built in this video is 1:4 impedance ratio. The primary has two turns through the core with a center tap and 4 turns on the secondary. That is a 1:2 turns ratio or a 1:4 impedance ratio (2 squared). You would need 4 more (8 total) turns on the secondary to get a 1:4 turns ratio or a 1:16 impedance ratio. The math 8/2 = 4 and 4 squared is 16. The other option is to have the primary make a single pass through each hole in the binocular core with the center tap on the far side of the core and then have 4 turns for the secondary. Anyhow, I would try more turns on the secondary and see where the output power lands. The plot thickens! I need more popcorn. de KB7ZUT

    • @na5y
      @na5y  3 года назад

      Thank you for the comment. Its hard to know what the schematic means by "1to16". I took it as 1:1:4 where each separate winding is 1/16th the impedance of the 4 turns. It could be 1:1:8 as you suggest. I just simmed 1:1:8 it in LTSpice and it didn't seem to make much of a difference interestingly to the final output waveform - it did increase the current through the IRF530 though. Gives me something else to play around with. Once again thank you for the comment and glad you enjoyed the series.

    • @na5y
      @na5y  3 года назад

      To add further confusion check out page 11 of this doc
      pa0fri.home.xs4all.nl/Lineairs/RF-AMP-2078/4Z4TJ%20(1).pdf
      Its for an earlier version of the board but this document says both four (in the diagram) *and* six turns (in the text).

    • @azav8raa
      @azav8raa 3 года назад

      The document you linked helps a little. So on page 11, it shows a 2:6 winding ration which is a 1:9 impedance ratio. And on page 8 they suggest even using a 2:7 winding ratio, which is a 1:12.25 impedance ratio. So based on that I would try adding 2-3 additional windings on the T3 secondary. Looks like they may have produced more Watts that way. Like I said, the "Plot Thickens." Half the fun of a no instructions kit is guessing what was really intended

    • @na5y
      @na5y  3 года назад +2

      @@azav8raa Heh - my thoughts exactly. If it was all clearly spelled out where would be the fun in that!.

  • @PabloWoiz
    @PabloWoiz Год назад

    Great videos!! R21 is missing, I putted a 100 Ohm resistor, and improves a lot the simetry of the output signal. Also it´s a mistake in the PCB in R20, a bridge is missing by factory.

    • @na5y
      @na5y  Год назад

      Interesting - never thought to play around with R21. Thank you for the comment

  • @remiechange5482
    @remiechange5482 2 года назад

    Thank you again for your work and the simulation files. I discover the very low input impedance of the second stage. I'm surprised.. Detail and suggestion: with 1 turn or 2, the coupling is not 1 but 0.95 or less. You can see it when you move the wire in the ferrite during the measurement. For a higher frequency, the R series in simulation must be added to L. Even more so with industry inductors.

    • @na5y
      @na5y  2 года назад

      Good suggestion on the coupling Remi - I had been wondering about that. Thanks again for the comment!

  • @9w2elm
    @9w2elm Год назад

    Great video sir, can I know how you adjust the 2 gate voltage and how many volts did you apply at each gate of the MOSFET?

    • @na5y
      @na5y  Год назад

      Unfortunately I don't recall the gate bias voltage - but have a look at my other comment. Just to confirm - you do have this mounted to a heatsink correct? Just thought I'd ask - the mosfets will blow very quickly if you don't have them on a good heatsink. Also make sure the mosfets are electrically isolated from the heatsink

    • @9w2elm
      @9w2elm Год назад

      @@na5y Thanks for the reply, I'm really appreciate your response. For your info, I mounted it on the heat sink and the MOSFET case was confirmed isolated from the heat sink it self

  • @9w2elm
    @9w2elm Год назад

    I'm putting 3.38V for the gate voltage but when I'm applying 14MHz -23dBm signal my current reading goes up and destroy the MOSFET, can you assists me regarding the problems?. For your info my first stage amplifier are ok and follow exactly measurements that your video show

    • @na5y
      @na5y  Год назад +1

      You will get variations in the IRF530s. What the schematic says is that the quiescent current thru the IRF530s should be 30mA each - adding to a total of 60mA.
      One way of accomplishing that is you first need a power supply that shows current draw. Then start with the bias turned all the way down on both mosfets - record the starting current draw. Now slowly turn RV1 up until you get an *additional* 30mA current draw. Once you do now turn RV2 up until you get an additional current draw of 30mA.
      At that point the mosfets should be biased correctly. Here's the schematic link that shows the 30mA
      drive.google.com/file/d/1rXdZkfNStUP0Zz3jFpwE6WsESj3K73Rh/view
      Hope this helps for you!

    • @9w2elm
      @9w2elm Год назад

      @@na5y For your info, I mounted it with a heat sink that I thought sufficient enough to maintain the MOSFET temperature. I'm also not monitored the drain current, I'm adjusting only the gate voltage to 3.5V. Also my power supply equipped with the analogue Ammeter. My output wave form are clean without any harmonics. Some how the current was keep rising when I monitored the output current.

  • @SeAfasia
    @SeAfasia 3 года назад

    Hi Ian,
    nice presentation....look some distortion at sine waves....but waiting for final tests...check the ferrites turns...thanks..

    • @na5y
      @na5y  3 года назад +1

      Yes - I am going to continue to play around with this a bit - thank you for the comment.

  • @R2AUK
    @R2AUK 3 года назад

    That was an interesting experiment, many thanks for sharing. Yeah, there are easier ways to get 10-15 W using a pair of cheap MOSFET's, e.g. IRF510's. I tried a schematic used in uBITX transceiver some time ago and it works quite well up to 14 Mhz. The output power drops to 3-4 W on 28 MHz. The MOSFETS can be replaced with RD15, and NPNs can be replaced with something that has higher fT. I tried KT3142A which are equal to 2N2369A and have fT = 500 Mhz. That gave me nice 5.5-7.5 W across all HF bands and the linearity was just great. I used this PA in my recent transceiver.

    • @na5y
      @na5y  3 года назад +1

      I just blew one of the mosfets seeing how far I could push the bias. I got up to 3.9V or so and then it went short circuit. Gives me a chance to put in a pair RD15s in there and see the difference

  • @lucaelectronic
    @lucaelectronic 2 года назад

    Hi,
    That's a nice project thanks for your good documentation about building the amplifier it's not that easy to find any information about it.

    • @na5y
      @na5y  2 года назад

      Thank you very much. I am glad you found it helpful!

  • @DAVIDGREGORYKERR
    @DAVIDGREGORYKERR 3 года назад

    That looks like the output amp of a COBRA 200 GTL DX Transceivers but your not using ceramic packaged MOSFETS.

  • @amirb715
    @amirb715 3 года назад

    did you check where in the circuit the distortion is introduced? is it Q2 or only the output stage? With so much distortion, can this comply with FCC rules for amateur radio?

    • @na5y
      @na5y  3 года назад

      Actually the distortion was significantly improved when I added cap C4 in. The schematic had it as not connected but there is a 270pF silver mica cap in the kit which goes there. In any case the amp certainly requires a low pass filter to be legal - I plan to do that next. Thank you for the comment Amir

    • @amirb715
      @amirb715 3 года назад

      @@na5y that makes sense. C4 plus the other two in parallel with the inductance of the transformer are probably shorting out (not quite completely) the higher harmonics. The loaded Q of that resonator is very low so lower harmonics like 2nd order is still going through perhaps but it helps to reduce the higher ones.

  • @mariovano
    @mariovano 3 года назад

    Well, I blew up one of the FETs, but I think it was my own fault. I assumed that the transformer driving them (T2) was one turn on the bipolar side and 4 turns on the FET side, I think that resulted in too high a voltage on the gates. Can you confirm which side of T2 is which? I've got plenty more fets on the way! Thanks again for doing this.
    AE0GL

    • @na5y
      @na5y  3 года назад

      Its 4 turns on the 2SC1971 side to 1 turn on the FET side for T2 Mario

    • @mariovano
      @mariovano 3 года назад

      @@na5y Obvious, now that the driver is working properly! I must have been putting more than 100V on the gates! - thanks.

    • @mariovano
      @mariovano 3 года назад

      After replacing the fets, I get good agreement with your results....AE0GL

    • @na5y
      @na5y  3 года назад

      DId you end up installing C4? I found the waveforms much better looking with that silver mica cap installed there

  • @amirb715
    @amirb715 3 года назад

    I think there is enough information for you to just draw the AC and DC load lines on the output (Ic-Vce) plots given in the datasheet and choose the best quiescent current based on the power and voltage swing at the collector. I think 0.65V DC at emitter, is probably too low. Q2 will hit the bottom and cut off very quickly and the waveform will be distorted too much...

    • @na5y
      @na5y  3 года назад

      I have seen the mathematical treatment of a simple transistor amplifier with resistor as load many times. I have never run across that same analysis where the load is an inductor. At the quiescent point for an amp with a resistor load you generally set the voltage at the collector to be around 1/2 Vcc (and draw your load line).
      With an inductor as load the quiescent voltage at the collector is Vcc because the inductor acts as a short at DC. The amplifier then operates around that point (rather than 1/2 Vcc). I'm sure there's an analysis out there somewhere - do you know of a reference Amir - I'd love to close that gap in my knowlege. Thank you for the comment.

    • @amirb715
      @amirb715 3 года назад

      @@na5y VC being equal to Vcc is fine, with the RE you can draw the DC load line but what is more critical is the AC load line for which you only consider the AC resistances seen at collector and emitter. That determines the voltage swing of the transistor.

  • @DAVIDGREGORYKERR
    @DAVIDGREGORYKERR 3 года назад

    If done right you should be able to get 60W RF power, You need to mount this on a heatsink before you burn out the MOSFETS.

    • @na5y
      @na5y  3 года назад

      Perhaps with a 20V+ supply you could achieve that - I don't think you could with this amplifer and just 12V (and remain linear). I could be wrong of course I am far from an expert in these things

    • @DAVIDGREGORYKERR
      @DAVIDGREGORYKERR 3 года назад

      @@na5y The COBRA 200 GTL DX puts out 100W of RF on transmit and uses 13.8V the driver is a MOSFET also but in the COBRA 200GTL DX is prone to burning out maybe that should go with a VN66AF with a heatsink for the driver.

    • @na5y
      @na5y  3 года назад

      @@DAVIDGREGORYKERR Certainly other PA circuits can get 100W output on 13.8 input - my ICOM can - but thats a whole other level of sophistication in the drive circuitry. I'll definitely be doing some playing around with this though to see how far I can push it. Thank you for the comment!

    • @DAVIDGREGORYKERR
      @DAVIDGREGORYKERR 3 года назад

      @@na5y The COBRA 200 GTL DX is for radio hams only not people with a licence.